Answer:
J=36221 Kg.m/s
Explanation:
Impulse-Momentum Theorem
These two magnitudes are related in the following way. Suppose an object is moving at a certain speed [tex]v_1[/tex] and changes it to [tex]v_2[/tex]. The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by
[tex]p=mv[/tex]
The initial and final momentums are, respectively
[tex]p_1=mv_1,\ p_2=mv_2[/tex]
The change of momentum is
[tex]\Delta p=p_2-p_1=m(v_2-v_1)[/tex]
It is numerically equal to the Impulse J
[tex]J=\Delta p[/tex]
[tex]J=m(v_2-v_1)[/tex]
We are given
[tex]m=2498\ kg,\ v_1=17.1\ m/s,\ v_2=2.6\ m/s[/tex]
The impulse the car experiences during that time is
[tex]J=2498(2.6-17.1)=2498(-14.5)[/tex]
J=-36221 Kg.m/s
The magnitude of J is
J=36221 Kg.m/s